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Wisconsin Public Radio wants to duplicate a survey conducted in 2011 that found that 68% of adults living in Wisconsin felt that the country was going in the wrong direction. How many people would need to be surveyed for a 90% confidence interval to ensure the margin of error would be less than 3%? Be sure to show all your work and round appropriately

User Natsu
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1 Answer

4 votes

Answer:

655 people would need to be surveyed.

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
\pi, and a confidence level of
1-\alpha, we have the following confidence interval of proportions.


\pi \pm z\sqrt{(\pi(1-\pi))/(n)}

In which

z is the zscore that has a pvalue of
1 - (\alpha)/(2).

In this question, we have that:


\pi = 0.68

The margin of error is:


M = z\sqrt{(\pi(1-\pi))/(n)}

90% confidence level

So
\alpha = 0.1, z is the value of Z that has a pvalue of
1 - (0.1)/(2) = 0.95, so
Z = 1.645.

How many people would need to be surveyed for a 90% confidence interval to ensure the margin of error would be less than 3%?

We need to survey n adults.

n is found when M = 0.03. So


M = z\sqrt{(\pi(1-\pi))/(n)}


0.03 = 1.645\sqrt{(0.68*0.32)/(n)}


0.03√(n) = 1.645√(0.68*0.32)


√(n) = (1.645√(0.68*0.32))/(0.03)


(√(n))^(2) = ((1.645√(0.68*0.32))/(0.03))^(2)


n = 654.3

Rounding up

655 people would need to be surveyed.

User Agektmr
by
8.0k points
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